Evaluate the integral for all positive whole numbers n: $$ \int^1_0 \left(\log\frac{1}{x}\right)^n dx $$

Question

zarkon · Answer

n!

dumbcow · Answer

wow how did you get n! ? 

using substitution i seem to get a repeating integration by parts 
u = log(x) --> x = e^u
du = dx/x

\[ - \int\limits_{0}^{1}e^{u}*u^{n} du

ybarrap · Answer

for positive whole numbers n plus zero http://www.wolframalpha.com/input/?i=integrate+%28+log+1%2Fx+%29^n+dx+from+x%3D0+to+1

zarkon · Answer

integration by parts

zarkon · Answer

$$u=\left(\ln(1/x)\right)^n$$

$$du=n(\ln(1/x))^{n-1}\frac{1}{1/x}\frac{-1}{x^2}=-n(\ln(1/x))^{n-1}(1/x)$$

$$dv=dx$$

$$v=x$$

zarkon · Answer

forgot a dx

zarkon · Answer

$$du=n(\ln(1/x))^{n-1}\frac{1}{1/x}\frac{-1}{x^2}dx=-n(\ln(1/x))^{n-1}(1/x)dx$$

anonymous · Answer

You can also use substitution to show that it equals $$ \Gamma(n+1) $$

zarkon · Answer

then use the formula for parts....you will get the reduction formula

zarkon · Answer

both just as trivial